Explanation

(b)
Given $\left| \begin{matrix}
1 & a & b \\
1 & c & a \\
1 & b & c
\end{matrix} \right| = 0$
$\Rightarrow c^{2} - ab - a(c - a) + b(b - c) = 0$
$\Rightarrow a^{2} + b^{2} + c^{2} - ab - bc - ca = 0$
${\Rightarrow \frac{1}{2}\lbrack(a - b)\ ^{2} + (b - c)^{2} + (c - a)^{2}\rbrack = 0
}{\Rightarrow a = b = c}$
So,$\mathrm{\Delta}ABC\$is equilateral triangle.
$\therefore\angle A = 60{^\circ},\ \angle B = 60{^\circ},\ \angle C = 60{^\circ}$
$\sin^{2}{A + \sin^{2}{B + \sin^{2}{C = \sin^{2}{60{^\circ} + \sin^{2}{60{^\circ} + \sin^{2}{60{^\circ}}}}}}}$
$= \left( \frac{\sqrt{3}}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2} + \left( \frac{\sqrt{3}}{2} \right)^{2}$
$= 3 \times \frac{3}{4} = \frac{9}{4}$